Can any one me say if there is a simple proof of this claim which i can prove it by localization and no easy technique of nuclear spaces...

Let $f:X\rightarrow S$ be an open, surjective morphism of complex reduced spaces with constant fiber dimension $n$ (or universally open morphism with $n$-fibers between locally noetherian excellent without embedded componnet schemes on field of charact.0).
Let $G$ be a coherent sheaf on $X$.

Claim: ${\rm R}^{n}f_{*}G=0$ if and only if the support of $G$ satisfies:

${\rm Supp}(G)\cap X_{s}$ is nowhere dense in the fiber $X_{s}$.

Remark:
1) We can assume $X$ $n$-complete space(i.e there exist a smooth strongly $n$-convex exhaustion function on $X$) and then all cohomology group $H^{k}(X, F)$ vanish for all coherent sheaf $F$ and all $k>n$.

2) For a projection $f:S\times U\rightarrow S$ where $U$ is an open polydisque of $C^{n}$, the result is true for ${\rm R}^{n}f_{!}$. By right exactness of the $n$-higher direct image functor, we can localize and deduce from the projection case the general case.... But, it is no simple proof...

Yes, I'm sure that the claim is what i want and trivially, for $X$ Stein or Stein morphism we have nothing to prove: all the cohomology object vanish.

You can assume $X$ $n$-complete complex space and then $H^{n}(X, G)$ is generally not finite dimensional.

In the general proof, the absolute case is essential and is proved by mean of Serre duality.
The relative case is not so trivial and make fundamental use of free nuclear resolution and quasi-coherency in analytic geometry.

I'm confused, I don't see why Laurent's example fails. You made the claim about ``if and only if''. He provided an example of an open surjective morphism $f : X \to S$ and a coherent sheaf $M$ such that $R^i M = 0$ for all $i$ but $M$ does \emph{not} satisfy your condition on the support.
–
Karl SchwedeSep 12 '10 at 14:58

Excuse me, we must add in the hypothesis that the morphism is not Stein morphism or the cohomology group $H^{n}(X, F}$ dont vanish for all coherent sheaf $F$ on $X$... In fact, we avoid the situation for which the cohomology or the higher direct image are trivial...